Answer:
CH3COOH would be more concentrated
Explanation:
The higher the concentration value, the more concentrated it is.
The relationship between concentration, moles and volume is given by the equation;
Concentration = No of moles / Volume
5.0 grams of HCOOH dissolved in 189 mL of water
Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol
Concentration = 0.1086 / 0.189 = 0.5746 mol/L
1.5 moles of CH3COOH dissolved in twice as much water
Volume = 2 * 189 = 378 ml = 0.378 L
Concentration = 1.5 / 0.378 = 3.9683 mol/L
Comparing both concentration values;
CH3COOH would be more concentrated
It would repel some of the negatively charged electrons of the wall so the answer would be C. The wall and the ballon repel each other
2.26%, 26.8%, 2 2/5, 2.62, 271%
Hope this helps! :D
The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂<span>O(l)
</span>
n=mass in g/M.M
15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂
no. of mol of HCl:
n=0.5 mol/L*0.075L=0.0375 mol
This could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.
Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.
Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂
no. of mol of CaCl₂= 0.0375/2= 0.01875 mol
mass in g=n*MM= 0.01875*111= 2.08 g
First calculate the moles of N2 and H2 reacted.
moles N2 = 27.7 g / (28 g/mol) = 0.9893 mol
moles H2 = 4.45 g / (2 g/mol) = 2.225 mol
We can see that N2 is the limiting reactant, therefore we
base our calculation from that.
Calculating for mass of N2H4 formed:
mass N2H4 = 0.9893 mol N2 * (1 mole N2H4 / 1 mole N2) * 32
g / mol * 0.775
<span>mass N2H4 = 24.53 grams</span>
